2000 AMC 12 Problems/Problem 17

Revision as of 19:44, 16 February 2010 by Poincare (talk | contribs) (Solution)

Problem

2000 12 AMC-17.png

A circle centered at $O$ has radius $1$ and contains the point $A$. The segment $AB$ is tangent to the circle at $A$ and $\angle AOB = \theta$. If point $C$ lies on $\overline{OA}$ and $\overline{BC}$ bisects $\angle ABO$, then $OC =$

$\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta}{1 + \sin \theta}\qquad \text {(D)}\ \frac{1}{1+\sin\theta} \qquad \text {(E)}\ \frac{\sin \theta}{\cos^2 \theta}$

Solution

Since $\overline{AB}$ is tangent to the circle, $\triangle OAB$ is a right triangle. Thus since $OA = 1$, $BA = \tan \theta$ and $OB = \sec \theta$. By the Angle Bisector Theorem, $\frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta$. Multiply both sides by $\cos \theta$ to simplify the trigonometric functions. Since $AC + OC = 1$, $1 - OC = OC \sin \theta (from$AC \sec \theta = OC \tan \theta$) \Longrightarrow$ $OC = \frac{1}{1+\sin \theta} \Rightarrow \mathrm{(D)}$.

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 12 Problems and Solutions